Quantum Science Philippines
Quantum Science Philippines

Radial Wavefunction of a Hydrogen Atom

Gibson T. Maglasang and John Paul Aseniero

In this article, we outlined the necessary steps in calculating the radial wavefunctions [eq]R_{nl}[/eq] for the Hydrogen atom. Thus, the radial wavefunctions particularly [eq]R_{30 [/eq], [eq]R_{31 [/eq] and [eq]R_{32 [/eq] are easily obtained  without bothering to normalize it.

We use the formula below to find the wavefunction,

[eq]R_{nl}=\frac{1}{r}\rho^{l+1}e^{-\rho}\nu(\rho),[/eq]                                ( 1)

where

[eq]\nu(\rho)=\sum_{j=0}^{\infty}c_j\rho^j,[/eq]                                         (2)

while [eq]c_j[/eq] is determined by the recursion formula given by,

[eq]c_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}c_j,[/eq]                          (3)

and

[eq]\rho=\frac{r}{na}.[/eq]                                                            (4)

(i) Now, finding [eq]R_{30}[/eq]

Using equation 1, we need to solve first the coefficient [eq]c_j[/eq] from (eqn. 3), with [eq]n=3[/eq] and [eq]l=0[/eq].

[eq]c_1=\frac{2(0+1-3)}{1(0+0+2)}c_0=-2c_0[/eq]

[eq]c_2=\frac{2(1+1-3)}{2(1+0+2)}c_1=-\frac{2}{3}c_0[/eq]

[eq]c_3=\frac{2(2+1-3)}{3(2+2)}c_2=0.[/eq]

Knowing the value of [eq]c_j[/eq], (eqn. 2) can now be easily determined,

[eq]\nu(\rho)=c_0\rho^0+c_1\rho^1+c_2\rho^2+c_3\rho^3.[/eq]                         (5)

Substituting the value of the calculated coefficients to (eqn. 5), we then have

[eq]\nu(\rho)=c_0-2c_0+\frac{2}{3}\rho^2c_0.[/eq]                                         (6)

Thus,

[eq]R_{30}(\nu)=\frac{1}{r}\Big(\frac{r}{3a}\Big)e^{-\rho}[c_0-2c_0+\frac{2}{3}c_0\rho^2].[/eq]              (7)

Plugging in (eqn. 4) to (eqn. 7), we finally have

[eq]R_30=\frac{c_0}{3a}\bigg[1-2\Big(\frac{r}{3a}\Big)+\frac{2}{3}\Big(\frac{r}{3a}\Big)\rho^2\bigg]e^{-(r/3a)}.[/eq]

Following the same process in (i), the rest of the wavefunctions are just straightforward.

(ii) For [eq]R_{31}[/eq]

Determining first the coefficients, with [eq]n=3[/eq] and [eq]l=1[/eq],

[eq]c_1=\frac{2(1+1-3)}{1(0+2+2)}c_0=-\frac{1}{2}c_0[/eq]

[eq]c_2=\frac{2(1+1+1-3)}{2(1+2+2)}c_0=0[/eq]

[eq]c_3=0[/eq]

Then,

[eq]\nu(\rho)=c_0-\frac{1}{2}c_0\rho[/eq].

Thus,

[eq]R_{31}=\Big(\frac{r}{9a^2}\Big)\bigg[1-\frac{r}{6a}\bigg]e^{-r/3a}[/eq]

(iii) We have the coefficients for [eq]R_{32}[/eq] with [eq]n=3[/eq] and [eq]l=2[/eq],

[eq]c_1=\frac{2(2+1-3)}{1(4+2)}c_0=0[/eq]

[eq]c_2=0[/eq]

[eq]c_3=0[/eq]

Using again (eqn. 1), we have

[eq]R_{32}=\frac{r^2}{27a^3}e^{-r/2a}c_0[/eq]

We finally generated the radial wavefunctions ([eq]R_{30}[/eq], [eq]R_{31}[/eq], [eq]R_{32}[/eq]) for the hydrogen atom which is the main aim of this paper.

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Finding the Expectation value for the ground state of a Hydrogen atom

John Paul Aseniero and Gibson T. Maglasang

For the particle in the state [eq]\Psi[/eq], the expectation value of x is expressed as

[eq]\langle x\rangle = \int_{-\infty}^{+\infty}x|\Psi(x,t)|^2dx[/eq]

where the expectation value is the average of repeated measurements on an ensemble of identically prepared systems.

In this article, we would like to find [eq]\langle r\rangle[/eq], [eq]\langle r^2\rangle[/eq], [eq]\langle x\rangle[/eq] and [eq]\langle x^2\rangle[/eq] for an electron in the ground state of a hydrogen atom and express it in Bohr radius.

a.) Calculating for the [eq]\langle r\rangle[/eq] and [eq]\langle r^2\rangle[/eq],

(i) Finding [eq]\langle r\rangle[/eq]

The ground state wavefunction for the Hydrogen atom is given by

[eq]\Psi_{100}(r\theta,\phi)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}[/eq]

Now getting the expectation value of r, we have

[eq]\langle r\rangle = \int{r|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]

[eq]=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}r^3 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]

[eq]=\int_0^{\infty}r^3 e^{-2r/a}dr[/eq]

The above integration can now  easily be facilitated by using the table of integral,

[eq]\int_0^{\infty}r^3 e^{-2r/a}dr = 3!(\frac{a}{2})^4[/eq].                   (1)

Therefore,

[eq] \langle r\rangle = \Big(\frac{4}{a^3}\Big)3!\Big(\frac{a}{2}\Big)^4[/eq].

(ii) For [eq]\langle r^2\rangle[/eq]

Next is we find the value of [eq]\langle r^2\rangle[/eq] by using the same process employed in the previous exercise. We have,

[eq]\langle r^2\rangle = \int r^2 |\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2(r^2\sin\theta d\theta d\phi dr)[/eq]

[eq]= \int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}r^4\frac{1}{\pi a^3}e^{-2r/a}\sin\theta d\theta d\phi dr[/eq]

[eq]=\frac{4}{a^3}\int_0^{\infty}r^4e^{-2r/a}dr[/eq]

Using again the table of integral used in (i) given in equation 1 to facilitate the integration, we get

[eq]=\frac{4}{a^3}4!\Big(\frac{a}{5}\Big)^5[/eq].

Thus,

[eq]\langle r^2\rangle=3a^2[/eq].

b) In the case of [eq]\langle x\rangle[/eq] and [eq]\langle x^2\rangle[/eq], for electron in ground state of hydrogen atom, this requires no new integration since [eq]r^2=x^2 + y^2 + z^2 [/eq].

(i) For the calculation of [eq]\langle x\rangle[/eq]

Now we have,

[eq]\langle x\rangle = \int{x|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]

but [eq] x= r\sin\theta \cos\phi [/eq], it implies that

[eq]\langle x\rangle=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)r^2 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]

Try to have a closer look at the integral of the [eq]\phi[/eq] part and evaluate it from 0 to [eq]2\pi[/eq]. Obviously we have,

[eq]\int_0^{2\pi}\cos\phi d\phi=[sin\phi]|_0^{2\pi}=0[/eq].

Therefore,

[eq]\langle x\rangle = 0[/eq].

(ii) However, for the [eq]\langle x^2\rangle[/eq]

To find for  [eq]\langle x^2\rangle[/eq], we have the following calculation,

[eq]\langle x^2\rangle = \int{x^2|\frac{1}{\sqrt{\pi a^3}}e^{-r/a}|^2}(r^2\sin\theta d\theta d\phi dr)[/eq]

[eq]=\frac{1}{\pi a^3}\int_{0}^{\infty}\int_0^{2\pi}\int_0^{\pi}(r\sin\theta\cos\phi)^2r^2 e^{-2r/a} \sin\theta d\theta d\phi dr[/eq]

[eq]= \frac{1}{\pi a^3}\int r^4e^{-2r/a}\sin^3\theta\cos^2\phi d\theta d\phi dr[/eq]

[eq]= \frac{1}{\pi a^3}\int_0^{\infty}r^4e^{-2r/a}\int_0^{2\pi}\cos^2\phi d\phi\int_0^{2\pi}\sin^3\theta d\theta[/eq]

Note that the above integration is facilitated by the following integral formulas:

[eq]\int_0^{2\pi} \sin^3\theta = \frac{4}{3}[/eq],

[eq]\int_0^{2\pi} \cos^2\phi d\phi = \pi[/eq].

Therefore,

[eq]\langle x^2\rangle = \frac{1}{\pi a^3}\frac{4}{3}\pi \int_{0}^{\infty} r^4 e^{-2r/a}[/eq]

[eq]= \frac{1}{3}\frac{4}{a^3}\int_0^{\infty} r^4 e^{-2r/a}dr[/eq]

[eq]=\frac{1}{3}\langle r^2\rangle[/eq].

So,

[eq]\langle x^2\rangle=a^2[/eq].

The expectation value for a ground state hydrogen atom are explicitly shown in this paper. The readers are also enjoined to calculate for the expectation value for momentum and see how they compare and contrast.

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Addition of Spin Angular Momentum

Eric Alcantara, Carlo Paul Morente and Gibson T. Maglasang

If you have two particles of spin [eq]S_1[/eq] and [eq]S_2[/eq]. Let [eq]\bf{S}[/eq] be the combined spin of the particles. You can get values of [eq]\bf{S}[/eq] from [eq](S_1+S_2)[/eq] down to [eq](S_1-S_2)[/eq]:

[eq]\bf{S} =(S_1+S_2), (S_1+S_2-1), (S_1+S_2-2),\cdot\cdot\cdot(S_1-S_2)[/eq]                (1)

We can get then the combination of states from the total spin state [eq]\bf{S}[/eq]. In particular, a state [eq]|S\quad m\rangle[/eq] with total S and z-component m will be some linear combination of the composite states [eq]|S_1\quad m\rangle[/eq] [eq]|S_2\quad m\rangle[/eq].

Consider now two spin 3/2 particles at the ground state. To find the possible total spin states, we use (eqn. 1). Thus,

[eq]\bf{S} = (3/2+3/2)= 3 ;[/eq]

[eq]\bf{S} = (3/2+3/2-1)= 2;[/eq]

[eq]\bf{S}=(3/2+3/2-2)= 1;[/eq]

[eq]\bf{S} = (3/2+3/2-3) = 0[/eq].

(i) For [eq]\bf{S} = 3[/eq], the corresponding values for [eq]m_s[/eq] are 3, 2, 1, 0, -1, -2, -3.

We have the following states [eq]|\bf{S}[/eq]  [eq] m_s\rangle[/eq],

[eq]|3[/eq]  [eq] 3\rangle[/eq],

[eq]|3[/eq]  [eq] 2\rangle[/eq],

[eq]|3[/eq]  [eq]1\rangle[/eq],

[eq]|3[/eq]  [eq]0\rangle[/eq],

[eq]|3[/eq]  [eq]-3\rangle[/eq],

[eq]|3[/eq]  [eq]-2\rangle[/eq],

[eq]|3[/eq]  [eq]-1\rangle[/eq].

(ii) For [eq]\bf{S}[/eq] = 2,   [eq]m_s=2, 1, 0, -1, -2[/eq]. The following states are obtained,

[eq]|2[/eq]  [eq] 2\rangle[/eq],

[eq]|2[/eq]  [eq] 1\rangle[/eq],

[eq]|2[/eq]  [eq] 0\rangle[/eq],

[eq]|2[/eq]  [eq] -2\rangle[/eq],

[eq]|2[/eq]  [eq] -1\rangle[/eq].

(iii) For [eq]\bf{S} = 1[/eq],    [eq]m_s = 1, 0, -1[/eq]

[eq]|1[/eq]  [eq]1\rangle[/eq],

[eq]|1[/eq]  [eq]0\rangle[/eq],

[eq]|1[/eq]  [eq]-1\rangle[/eq].

(iv) For [eq]\bf{S} = 0[/eq], [eq]m_s = 0[/eq]. Only one state is available for [eq]\bf{S} = 0[/eq]. We have,

[eq]|0[/eq]  [eq]0\rangle[/eq]

Thus, we have successfully extracted the 16 possible spin states for two spin 3/2 particles at the ground state.  One can also get all the 16 spin states for this particular problem by looking up the Clebsch-Gordan table.

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Addition of Angular Momenta

Sandra Manulat and Michael Jabines

Consider two spin 3/2 particles at the ground state. Suppose that their total spin is 2 with its z-component equal to zero. If you measure [eq]J_z[/eq] on the second particle, what values would you get and the probabilities of measuring such?

In order to answer the problem, we first need to learn the addition of angular momenta.

Let us define the total angular momentum operator:

[eq]\vec{J} =\vec{ J}_1 + \vec{J}_2 [/eq]

where [eq]\vec{J}_1[/eq] can be the orbital angular momentum operator and [eq]\vec{ J}_2[/eq] is the spin angular momentum operator.

Moreover, [eq]\vec{J_1}[/eq] and [eq]\vec{J_2}[/eq] can also denote the angular momentum operators of two different particles in a multi-particle system just like the problem above.

By definition, angular momentum operators are Hermitian, hence [eq]J[/eq], which is a just the sum of two Hermitian operators is also Hermitian satisfying the commutation relation

[eq]\vec{ J} \times \vec{J} = i\hbar\vec{ J} [/eq]

Thus, [eq]\vec{J}[/eq] possesses the expected properties of an angular momentum operator:

[eq] J^2 = j(j + 1)\hbar^2 [/eq]

[eq] J_z = m\hbar^2 [/eq]

where [eq]m = j, j-1, \cdots, -j+1, -j[/eq]

Now, we find a relationship between the quantum number of the total angular momentum, [eq]j[/eq] and [eq]m[/eq] with the individual quantum numbers [eq]j_1[/eq], [eq]j_2[/eq], [eq]m_1[/eq], and [eq]m_2[/eq].

Recall that,

[eq][J^2, J_1] = [J^2, J_2] = 0[/eq]

this implies that we can solve for the quantum numbers [eq]j[/eq], [eq]j_1[/eq], and [eq]j_2[/eq] simultaneously.  On the other hand,

[eq][J^2, J_{z1}] \neq 0[/eq]

[eq][J^2, J_{z2}] \neq 0[/eq]

which tells us that the quantum numbers [eq]m_1[/eq] and [eq]m_2[/eq] cannot be solved simultaneously with the quantum number [eq]j[/eq].

What to do?

Now we need to form two alternate groups of mutually commuting operators.

1st group: the eigenkets of [eq]{J_1}^2, {J_2}^2, J_{z1}[/eq] and [eq]J_{z2}[/eq] are [eq]|j_1, j_2; m_1, m_2>[/eq]

2nd group: the eigenkets of [eq]{J_1}^2, {J_2}^2, J, J_z[/eq] are [eq]|j_1, j_2; j, m>[/eq]

Each set of eigenkets are complete, mutually orthogonal and have unit norms, then we can write a conventional completeness relation,

[eq]\sum_{m_1}\sum_{m_2} |j_1, j_2; m_1, m_2><j_1, j_2; m_1, m_2 | = 1[/eq]

[eq]\sum_{j}\sum_{m} |j_1, j_2; j, m><j_1, j_2; j, m | = 1[/eq]

The summation is over all allowed values of [eq]m_1, m_2, j[/eq] and [eq]m[/eq].

These relation tells us that if we know [eq]j_1, j_2, j, m[/eq] we will be able to solve for a range of possible values of  [eq]m_1[/eq] and [eq]m_2[/eq] .

Now we ask: Given a state [eq]|j_1, j_2 ; m_1, m_2>[/eq], how do we express it in terms of [eq]|j_1, j_2; j, m>[/eq]?

It’s simple, we can use the completeness relation above to write,

[eq]|j_1, j_2; j, m> = \sum_{m_1}\sum_{m_2}<j_1, j_2; m_1, m_2|j_1, j_2; j, m>|j_1, j_2; m_1, m_2>[/eq]

Writing it more simply we have,

[eq]|j,m> = \sum_{m_1,m_2} C_{j_1, j_2; m_1, m_2}^{j, m} |j_1, m_1>|j_2, m_2>[/eq]

where [eq]C_{j_1, j_2; m_1, m_2}^{j, m}\;\;\;\; =\;\;\;\; \langle j_1, j_2; m_1, m_2 | j_1, j_2; j, m\rangle [/eq] is called the Clebsch-Gordan coefficients.

At this point, we are ready to answer our problem. We don’t need to worry solving for the Clebsch-Gordan coefficients because a table is made ready for our use.

Given: two spin 3/2 particles:

[eq]j_1= \frac{3}{2},\;\;\;\; j_2 =\frac{3}{2}[/eq]

[eq]j \;\;\;\; = \;\;\;\; 2, \;\;\;\; m\;\;\;\;=\;\;\;\; 0[/eq]

[eq]m = m_1 + m_2 = 0[/eq]

Now [eq]m_1 [/eq] and [eq]m_2[/eq] can have values in the range [eq]j_1, j_1 – 1, \cdots, -j_1 + 1, -j_1[/eq], and [eq]j_2, j_2 – 1, \cdots, -j_2 + 1, -j_2[/eq] respectively.

Thus,

[eq]|2\;\;\;\;0>\;\;\;\; = \;\;\;\;C_1 |2\;\;\;\;\frac{3}{2}>|0\;\;\;\;-\frac{3}{2}>\;\; +\;\; C_2 |2\;\;\;\;\frac{1}{2}>|0\;\;\;\;-\frac{1}{2}>[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + C_3 |2\;\;\;\;-\frac{1}{2}>|0\;\;\;\;\frac{1}{2}>\;\; +\;\; C_4 |2\;\;\;\;-\frac{3}{2}>|0\;\;\;\;\frac{3}{2}>[/eq]

For the values of [eq]C_1, C_2, C_3, C_4[/eq] we’ll use the table of Clebsch-Gordan coefficients, specifically the 3/2 by 3/2 table since we have two spin 3/2 particles:

Clebsch-Gordan coefficients for spin 3/2 two particle system. The shaded region corresponds to the probability of the corresponding spin state.

Clebsch-Gordan coefficients for spin 3/2 two particle system. The shaded region corresponds to the probability of the corresponding spin state.

Now we have,

[eq]|2\;\;\;\;0>\;\;\;\; = \;\;\;\;\frac{1}{\sqrt{4}} |2\;\;\;\;\frac{3}{2}>|0\;\;\;\;-\frac{3}{2}>\;\; +\;\; \frac{1}{\sqrt{4}} |2\;\;\;\;\frac{1}{2}>|0\;\;\;\;-\frac{1}{2}>[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -\frac{1}{\sqrt{4}} |2\;\;\;\;-\frac{1}{2}>|0\;\;\;\;\frac{1}{2}>\;\; \;\;-\frac{1}{\sqrt{4}} |2\;\;\;\;-\frac{3}{2}>|0\;\;\;\;\frac{3}{2}>[/eq]

Finally, we are told to measure the z-component of the spin on the second particle: [eq]J_{z2} = m_2\hbar[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;J_{z2} \;\;\;\;\;\;\;\;\;\;Probability[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{3}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;-\frac{3}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;-\frac{1}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

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Quantum Mechanics in Three Dimension: Angular Momentum and Probabilities

Hananish Joy Odarve and Edmar Pantohan

In this article, we want to show that given the state occupied by an electron in a hydrogen atom, one can measure the operators [eq]L^2[/eq] , [eq]S^2[/eq] , [eq]L_z[/eq], [eq]S_z[/eq], [eq]J^2[/eq] and [eq]J_z[/eq] and its corresponding probabilities.

Lets look at an example to get a clearer picture of these topic.  The electron in a hydrogen atom occupies the combined spin and position state given by

[eq]\Psi = R_{21} \;\; \left[\sqrt{\frac{1}{3}} \;\; {Y^0_1} \;\; \chi_+ \;\;\;\; + \;\;\;\; \sqrt{\frac{2}{3}} \;\; {Y^1_1} \;\; \chi_- \right] [/eq]

FOR [eq]L^2[/eq]:

First, we measure the orbital angular momentum squared given by

[eq]L^2 = l(l+1) {\hbar}^2 [/eq]

From the given state, we consider [eq]{Y^0_1}[/eq] where [eq]l=1[/eq] and [eq]m=o[/eq], so

[eq]L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2[/eq]

and

[eq]P = \frac{1}{3}[/eq].

Next, we consider [eq]{Y^1_1}[/eq] where [eq]l=1[/eq] and [eq]m=1[/eq], so

[eq]L^2 =1(1 + 1)\hbar}^2 \\= 2{\hbar}^2[/eq]

and

[eq]P = \frac{2}{3}[/eq].

Therefore, for the state given, the value of [eq]L^2 = 2{\hbar}^2[/eq] and the total probability is:

[eq]P =\frac{1}{3} + \frac{2}{3} = 1[/eq].

FOR [eq]L_z[/eq]:

Now, we measure the [eq]z[/eq] component of the orbital angular momentum which is given by

[eq]L_z = m\hbar[/eq].

So, for [eq]{Y^0_1}[/eq] where [eq]m=o[/eq]

[eq]L_z = 0\hbar =0[/eq] with [eq]P = \frac{1}{3}[/eq]

and for [eq]{Y^1_1}[/eq] where [eq]m=1[/eq]

[eq]L_z = 1\hbar =\hbar[/eq] with [eq]P = \frac{2}{3}[/eq].

FOR [eq]S^2[/eq]:

Next, we measure the spin angular momentum squared given by

[eq]S^2 = s(s+1) {\hbar}^2 [/eq]

Since [eq]s=\frac{1}{2}[/eq], there is only one possible value and this is

[eq]S^2 = \frac{1}{2}\left(\frac{1}{2}+1\right) {\hbar}^2 \\ = \frac{1}{2}\left(\frac{3}{2}\right){\hbar}^2 \\ = \frac{3}{4}{\hbar}^2[/eq]

and the probabilty then is [eq]P = 1[/eq].

FOR [eq]S_z[/eq]:

Here, we measure the [eq]z[/eq] component of the spin angular momentum which is given by

[eq]S_z = m_s\hbar [/eq].

For [eq]s=\frac{1}{2}[/eq], the value of [eq]m_s=\pm \frac{1}{2}[/eq] so for [eq]m_s=+\frac{1}{2}[/eq]:

[eq]S_z = +\frac{1}{2}\hbar [/eq]  with  [eq]P = \frac{1}{3}[/eq]

and for [eq]m_s=-\frac{1}{2}[/eq]:

[eq]S_z = -\frac{1}{2}\hbar [/eq]  with  [eq]P = \frac{2}{3}[/eq].

FOR [eq]J^2[/eq]:

Now, to calculate for the total angular momentum, we rewrite the given state of the system as:

[eq]\sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \left|10\right \rangle + \sqrt{\frac{2}{3}}|\frac{1}{2}\;\;-\frac{1}{2}\rangle \left|11\right \rangle[/eq]

This is for particles of spin 1 and spin [eq]\frac{1}{2}[/eq] so we use Clebsch-Gordan table for [eq] 1\times\frac{1}{2}[/eq] and so we have:

= [eq]\sqrt{\frac{1}{3}}\;\;\left[\sqrt{\frac{2}{3}}|\frac{3}{2}\;\;\frac{1}{2}\rangle – \sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right ][/eq] + [eq]\sqrt{\frac{2}{3}}\;\;\left[\sqrt{\frac{1}{3}}|\frac{3}{2}\;\;\frac{1}{2}\rangle + \sqrt{\frac{2}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right ][/eq]

= [eq]\left(2\frac{\sqrt{2}}{3}\right) |\frac{3}{2}\;\;\frac{1}{2}\rangle [/eq] + [eq]\left(\frac{1}{3}\right) |\frac{1}{2}\;\;\frac{1}{2}\rangle [/eq]

from here, we can say that [eq]j=\frac{3}{2}[/eq] or [eq]\frac{1}{2}[/eq] and so for [eq]j=\frac{3}{2}[/eq]:

[eq]J^2 = \frac{3}{2}\left(\frac{3}{2}+1\right) {\hbar}^2 =\frac{3}{2}\left(\frac{5}{2}\right){\hbar}^2 =\frac{15}{4}{\hbar}^2[/eq]

with

[eq]P = \frac{1}{3}[/eq].

And for [eq]j=\frac{1}{2}[/eq]:

[eq]J^2 = \frac{1}{2}\left(\frac{1}{2}+1\right) {\hbar}^2 =\frac{1}{2}\left(\frac{3}{2}\right){\hbar}^2 =\frac{3}{4}{\hbar}^2[/eq]

with

[eq]P = \frac{2}{3}[/eq].

FOR [eq]J_z[/eq]:

Now, we measure the [eq]z[/eq] component of the total angular momentum which is given by

[eq]J_z = m_j\hbar [/eq]

and

[eq]m_j= m+m_s[/eq].

for [eq]{Y^0_1} \;\; \chi_+[/eq], [eq]m=0[/eq] and [eq]m_s=\frac{1}{2}[/eq], [eq]m_j=\frac{1}{2}[/eq] and so

[eq]J_z = \frac{1}{2}\hbar [/eq].

for [eq]{Y^1_1} \;\; \chi_-[/eq], [eq]m=1[/eq] and [eq]m_s=-\frac{1}{2}[/eq], [eq]m_j=\frac{1}{2}[/eq] and so it is still

[eq]J_z = \frac{1}{2}\hbar [/eq].

Therefore, the probability of getting the value of [eq]J_z = \frac{1}{2}\hbar [/eq] is

[eq]P =1[/eq].

FOR the Probabilty density of finding the particle at [eq]\left(r, \theta, \phi\right )[/eq]:

[eq]\left|\Psi\right|^2 = \Psi*\Psi = {R_{21}}^* \left[\sqrt{\frac{1}{3}}{Y^0_1}^* \chi_+ + \sqrt{\frac{2}{3}}{Y^1_1}^* \chi_- \right] [/eq][eq]R_{21} \left[\sqrt{\frac{1}{3}}{Y^0_1} \chi_+ + \sqrt{\frac{2}{3}}{Y^1_1} \chi_- \right] [/eq]

[eq]= \left|R_{21}\right|^2\; \frac{1}{3}\left|Y^0_1\right|^2\left(\chi^\dagger_+\chi_+\right) [/eq]  + [eq]\left|R_{21}\;\right|^2\frac{\sqrt 2}{3} \left[{Y^0_1}^*{Y^1_1}\left(\chi^\dagger_+\chi_-\right) + {Y^1_1}^*{Y^0_1}\left(\chi^\dagger_-\chi_+\right) \right][/eq] + [eq]\left|R_{21}\;\right|^2 \frac{2}{3}\left|Y^1_1\right|^2\left(\chi^\dagger_-\chi_-\right)[/eq]

[eq]= \frac{1}{3}\left|R_{21}\right|^2 \left( \left|Y^0_1\right|^2 + 2\left|Y^1_1\right|^2\right)[/eq].

\rangle – \sqrt{\frac{1}{3}}|\frac{1}{2}\;\;\frac{1}{2}\rangle \right]

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